A344235 Triangle T from the array A(k, n) giving the sums of k+1 consecutive squares starting with n^2, read as upwards antidiagonals, for k >= 0 and n >= 0.
0, 1, 1, 5, 5, 4, 14, 14, 13, 9, 30, 30, 29, 25, 16, 55, 55, 54, 50, 41, 25, 91, 91, 90, 86, 77, 61, 36, 140, 140, 139, 135, 126, 110, 85, 49, 204, 204, 203, 199, 190, 174, 149, 113, 64, 285, 285, 284, 280, 271, 255, 230, 194, 145, 81, 385, 385, 384, 380, 371, 355, 330, 294, 245, 181, 100
Offset: 0
Examples
The array A(k, n) begins: k \ n 0 1 2 3 4 5 6 7 8 9 10 ... ----------------------------------------------------------- 0: 0 1 4 9 16 25 36 49 64 81 100 ... 1: 1 5 13 25 41 61 85 113 145 181 221 ... 2: 5 14 29 50 77 110 149 194 245 302 365 ... 3: 14 30 54 86 126 174 230 294 366 446 534 ... 4: 30 55 90 135 190 255 330 415 510 615 730 ... 5: 55 91 139 199 271 355 451 559 679 811 955 ... 6: 91 140 203 280 371 476 595 728 875 1036 1211 ... 7: 140 204 284 380 492 620 764 924 1100 1292 1500 ... 8: 204 285 384 501 636 789 960 1149 1356 1581 1824 ... 9: 285 385 505 645 805 985 1185 1405 1645 1905 2185 ... ... ----------------------------------------------------------- The triangle T(m, n) begins: m \ n 0 1 2 3 4 5 6 7 8 9 ... ----------------------------------------------------------- 0: 0 1: 1 1 2: 5 5 4 3: 14 14 13 9 4: 30 30 29 25 16 5: 55 55 54 50 41 25 6: 91 91 90 86 77 61 36 7: 140 140 139 135 126 110 85 49 8: 204 204 203 199 190 174 149 113 64 9: 285 285 284 280 271 255 230 194 145 81 ... ----------------------------------------------------------
References
- Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2nd ed.; Addison-Wesley, 1994, pp. 283-290.
Crossrefs
Formula
A(k, n) = Sum_{j=0..k} (n+j)^2, for k >= 0, n >= 0.
A(k, n) = Sum_{j=0..n+k} j^2 - (2*n-1)*n*(n-1)/3! = S(n+k) - (2*n-1)*n*(n-1)/3!, with S(n+k) = (1/3)*Sum_{j=0..2} binomial(3, j)*B_j*(n+k+1)^(3-j), with the Bernoulli numbers A027641 / A027642 (see Graham et al., pp. 283-290).
Recurrence for sequence of row k: A(k, n) = A(k, n-1) + (k+1)*(2*n + k - 1), n >= 1, with A(k, 0) = (2*k+1)*(k+1)*k/3!, for k >= 0.
Comments